A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is : A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.
A.16
B.14
C.10
D.12
Solution & Explanation
### Related Formula
For a wire of total resistance R$R$ divided into N$N$ equal segments, the resistance of each segment r$r$ is:
r = fracRN$$r = \frac{R}{N}$$
For a balanced Wheatstone bridge with resistors of resistance r$r$, the central arm can be neglected because no current flows through it.
### Core Logic
The triangular pyramid has 6$6$ segments of equal length. Since the total resistance of the wire is R$R$:
r = fracR6$$r = \frac{R}{6}$$
Let the four vertices of the pyramid be A$A$, B$B$, C$C$, and D$D$. Terminals are at A$A$ and B$B$.
The segments are:
- AB$AB$ (direct path between terminals, resistance r$r$)
- AC$AC$, BC$BC$, AD$AD$, BD$BD$ (forming a closed quadrilateral network between A$A$ and B$B$ with bridge arm CD$CD$)
- CD$CD$ (bridge arm connecting the midpoints, resistance r$r$)
### Step 1: Simplify the Network
By symmetry, the potentials at C$C$ and D$D$ are equal when a voltage is applied across A$A$ and B$B$. Thus, the bridge is balanced, and no current flows through the segment CD$CD$.
We can remove segment CD$CD$ from the calculations:
- The path A to C to B$A \to C \to B$ consists of two resistors in series: r + r = 2r$r + r = 2r$.
- The path A to D to B$A \to D \to B$ also consists of two resistors in series: r + r = 2r$r + r = 2r$.
- The direct path A to B$A \to B$ has a single resistor r$r$.
These three parallel branches are connected between A$A$ and B$B$.
### Step 2: Calculate Equivalent Resistance
The equivalent resistance R_AB$R_{AB}$ is:
frac1R_AB = frac12r + frac12r + frac1r = frac1r + frac1r = frac2r$$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$$R_AB = fracr2$$R_{AB} = \frac{r}{2}$$
Substitute r = fracR6$r = \frac{R}{6}$:
R_AB = fracR/62 = fracR12$$R_{AB} = \frac{R/6}{2} = \frac{R}{12}$$
Comparing with R_AB = fracRn$R_{AB} = \frac{R}{n}$, we find n = 12$n = 12$.
### Pattern Recognition
Sees: Resistor network formed by a 3D pyramid (6 identical edges, 4 nodes).
Shortcut: A 6-resistor regular tetrahedron has an equivalent resistance of r/2$r/2$ across any two vertices. Since the total wire resistance is R$R$ and it's cut into 6 pieces, r = R/6 implies R_texteq = R/12$r = R/6 \implies R_{\text{eq}} = R/12$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.
Keywords:#wire bent into triangular pyramid#JEE Main 2025 Morning Q3#Current Electricity JEE Main 2025#equivalent resistance Wheatstone bridge#resistor network#equivalent resistance#balanced wheatstone bridge#pyramid topology
More Current Electricity Previous-Year Questions — Page 3
Q6jee_main_2025_04_april_eveningCombination of Resistors
From the combination of resistors with resistance valuesR_1=R_2=R_3=5\ Omega$R_{1}=R_{2}=R_{3}=5\ \Omega$ and R_4=10\ Omega$R_{4}=10\ \Omega$, which of the following combination is the best circuit to get an equivalent resistance of 6\ Omega$6\ \Omega$?
A.
B.
C.
D.
Solution
### Related Formula
Series combination:
R_s = R_a + R_b$R_s = R_a + R_b$
Parallel combination:
frac1R_p = frac1R_1 + frac1R_2 implies R_p = fracR_1 R_2R_1 + R_2$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_p = \frac{R_1 R_2}{R_1 + R_2}$$
### Core Logic
Let's check option (1):
Top branch has R_1$R_1$ and R_2$R_2$ in series: R_texttop = 5 + 5 = 10\ Omega$R_{\text{top}} = 5 + 5 = 10\ \Omega$.
Bottom branch has R_3$R_3$ and R_4$R_4$ in series: R_textbottom = 5 + 10 = 15\ Omega$R_{\text{bottom}} = 5 + 10 = 15\ \Omega$.
### Step 1: Calculate Equivalent Parallel Resistance
These two branches are connected in parallel, so the total equivalent resistance is:
R_p = frac10 times 1510 + 15 = frac15025 = 6\ Omega$$R_p = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega$$
This matches the target value of 6\ Omega$6\ \Omega$ perfectly. Circuit diagram analysis for 6 ohm equivalent resistance
### Pattern Recognition
Look for symmetric partitions. Standard combinations of values like 10\ Omega$10\ \Omega$ and 15\ Omega$15\ \Omega$ yield exactly 6\ Omega$6\ \Omega$ in parallel.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q13jee_main_2025_04_april_morningElectric Current and Charge Flow
Current passing through a wire as function of time is given as I(t)=0.02t+0.01mathrm~A$I(t)=0.02t+0.01\mathrm{~A}$. The charge that will flow through the wire from t=1mathrm~s$t=1\mathrm{~s}$ to t=2mathrm~s$t=2\mathrm{~s}$ is:
### Core Logic
The total resistance of the continuous uniform wire loop is 9Omega$9\Omega$. When bent into an equilateral triangle, it is split into three equal length sections. The resistance of each individual side is:
R_textside = frac9Omega3 = 3Omega$$R_{\text{side}} = \frac{9\Omega}{3} = 3\Omega$$
### Step 1: Calculating Equivalent Series and Parallel Resistance
As shown in the circuit diagrams Combination of Resistors diagram for Q23 - JEE Main 2025 Morning and Combination of Resistors diagram for Q23 - JEE Main 2025 Morning, measuring across any two vertices means one branch contains a single side resistor (3Omega$3\Omega$), while the other branch contains the remaining two sides connected in series :
R_textseries = 3Omega + 3Omega = 6Omega$$R_{\text{series}} = 3\Omega + 3\Omega = 6\Omega$$
Now, calculate the parallel equivalent between these two branches:
R_texteq = fracR_textside times R_textseriesR_textside + R_textseries = frac3 times 63 + 6 = frac189 = 2Omega$$R_{\text{eq}} = \frac{R_{\text{side}} \times R_{\text{series}}}{R_{\text{side}} + R_{\text{series}}} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\Omega$$
### Pattern Recognition
For a closed uniform loop with N$N$ equal sides, the parallel resistance measured across adjacent corners always simplifies to fracN-1N^2 cdot R_texttotal$\frac{N-1}{N^2} \cdot R_{\text{total}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q24jee_main_2025_28_jan_eveningWheatstone Bridge
The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ________ A.
A bridge resistor network supplied by a 40V DC voltage source terminal layout.
Numerical Answer.Answer: 2
Solution
### Related Formula
For a balanced Wheatstone bridge network, if the potentials at opposite nodes are equal (V_A = V_B$V_A = V_B$), no current flows through the central branch. The resistance arms satisfy the balance ratio:
fracR_1R_2 = fracR_3R_4$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$
Total current from the source is calculated using Ohm's law:
I = fracV_textsourceR_textequivalent$$I = \frac{V_{\text{source}}}{R_{\text{equivalent}}}$$
### Core Logic
Given conditions :
V_A = V_B implies textBalanced condition$$V_A = V_B \implies \text{Balanced condition}$$
From the circuit network diagram [cite: 817, 818, 826, 828]:
* Left-top arm = 10 \ Omega$= 10 \ \Omega$
* Left-bottom arm = R$= R$
* Right-top arm = 20 \ Omega$= 20 \ \Omega$
* Right-bottom arm = 40 \ Omega$= 40 \ \Omega$
Apply the balance condition to solve for unknown resistor R$R$ :
frac10R = frac2040 implies frac10R = frac12 implies R = 20 \ Omega quad text[cite: 837, 838]$$\frac{10}{R} = \frac{20}{40} \implies \frac{10}{R} = \frac{1}{2} \implies R = 20 \ \Omega \quad \text{[cite: 837, 838]}$$
Now restructure the equivalent network :
Since the central 30 \ Omega$30 \ \Omega$ resistor branch carries zero current, it can be removed from the calculation [cite: 820, 834].
* Top \parallel branch: 10 \ Omega + 20 \ Omega = 30 \ Omega$10 \ \Omega + 20 \ \Omega = 30 \ \Omega$
* Bottom \parallel branch: 20 \ Omega + 40 \ Omega = 60 \ Omega$20 \ \Omega + 40 \ \Omega = 60 \ \Omega$
Calculate total equivalent resistance R_texteq$R_{\text{eq}}$:
R_texteq = frac30 times 6030 + 60 = frac180090 = 20 \ Omega$$R_{\text{eq}} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20 \ \Omega$$
Calculate total current I$I$ drawn from the 40textV$40\text{V}$ source:
I = frac40 text V20 \ Omega = 2 text A$$I = \frac{40 \text{ V}}{20 \ \Omega} = 2 \text{ A}$$
### Step 1: Circuit Solution
The balanced bridge network layout with branch currents is shown below:
A bridge resistor network supplied by a 40V DC voltage source terminal layout.
### Pattern Recognition
When a question states that two nodes are at equal potential (V_A = V_B$V_A = V_B$), immediately identify it as a balanced Wheatstone bridge. This allows you to remove the central branch and simplify the circuit into basic series-\parallel resistor combinations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q34jee_main_2024_01_february_morningCells and EMF
The reading in the ideal voltmeter (V) shown in the given circuit diagram is:
A schematic showing a symmetric loop containing multiple identical 5V sources with 0.2 Ohm internal resistances coupled across an ideal voltmeter terminal.
A.5mathrm~V$5\mathrm{~V}$
B.10mathrm~V$10\mathrm{~V}$
C.0mathrm~V$0\mathrm{~V}$
D.3mathrm~V$3\mathrm{~V}$
Solution
### Related Formula
Total effective loop EMF and internal resistance:
I = fracE_textnetr_textnet$$I = \frac{E_{\text{net}}}{r_{\text{net}}}$$
Terminal potential difference across a discharging cell:
V = E - Ir$V = E - Ir$
### Core Logic
The loop has a set of 8 cells all aiding the same current flow sequence direction.
E_texteq = 8 times 5 = 40mathrm~V$$E_{\text{eq}} = 8 \times 5 = 40\mathrm{~V}$$r_texteq = 8 times 0.2 = 1.6mathrm~Omega$$r_{\text{eq}} = 8 \times 0.2 = 1.6\mathrm{~\Omega}$$
The circulating loop current is:
I = frac40mathrm~V1.6mathrm~Omega = 25mathrm~A$$I = \frac{40\mathrm{~V}}{1.6\mathrm{~\Omega}} = 25\mathrm{~A}$$
### Step 1: Terminal Voltage Calculation
The ideal voltmeter measures the potential drop across the localized parallel branch configuration:
V = E - Ir = 5 - (25 times 0.2) = 5 - 5 = 0mathrm~V$$V = E - Ir = 5 - (25 \times 0.2) = 5 - 5 = 0\mathrm{~V}$$
### Pattern Recognition
A closed loop composed entirely of identical series active cells short circuits itself perfectly relative to the localized node drop points, reducing the net external voltage drop to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
More Current Electricity Questions — jee_main_2025_07_april_morning
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